Question

according to the general trend of electronegativity, which of the following elements has the smallest electronegativity: li, mg, na, or be?
○ mg
○ na
○ li
○ be
clear my selection
8 multiple choice 1 point
the shortest length separating the nuclei of the two atoms during a collision refers to...
○ nonbonding radius.
○ bonding atomic radius.
○ electron radius.
○ ion radius.
clear my selection

Explanation:

First Question (Electronegativity)

Step1: Recall electronegativity trends

Electronegativity increases across a period (left to right) and decreases down a group (top to bottom) in the periodic table.

Step2: Locate elements in the periodic table

• Li (lithium) and Na (sodium) are in Group 1 (alkali metals), with Na below Li (so Na has lower electronegativity than Li).
• Be (beryllium) and Mg (magnesium) are in Group 2 (alkaline earth metals), with Mg below Be (so Mg has lower electronegativity than Be).
• Comparing Group 1 and Group 2, Group 1 metals have lower electronegativity than Group 2 metals in the same period. Na is below Li (Group 1), so Na has the lowest electronegativity among Li, Mg, Na, Be.

Step1: Recall definitions of atomic radii

• Nonbonding radius: Distance between nuclei of two non - bonded atoms.
• Bonding atomic radius: Half the distance between the nuclei of two bonded atoms (or the shortest distance between nuclei during collision as atoms approach and then repel; this is the bonding atomic radius concept).
• Electron radius: Not a standard term for this context.
• Ion radius: Radius of an ion, not related to atomic collision distance.

Step2: Match the definition

The description "the shortest length separating the nuclei of the two atoms during a collision" matches the definition of the bonding atomic radius.

Answer:

B. Na

Second Question (Atomic Radius Definition)